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From: WM <wolfgang.mueckenheim@tha.de>
Newsgroups: sci.math
Subject: Re: Incompleteness of Cantor's enumeration of the rational numbers
Date: Tue, 26 Nov 2024 10:11:22 +0100
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On 25.11.2024 22:05, joes wrote:
> Am Mon, 25 Nov 2024 13:18:28 +0100 schrieb WM:

>>> But there is no finite set with ALL natural numbers.
>>> Like usual, you mess up with your qualifiers.
>> ℕ is fixed, that means |ℕ| is fixed.
> What does that have to do with it?

It is impossible to add or to delete an element.
It is impossible to change |ℕ| by 1 or more.
> 
>>>>> Limit theory only works if the limit actually exists
>>>> If limits exist at all, then the limit of the sequence 1/10, 1/10,
>>>> 1/10, ... does exist.
>>> But the concept of 1/10th of an infinte set does not exist..
>> It does.
> It has the same cardinality.

Yes, it is much.
> 
>>>>> You can get things that APPEAR to reach a limit, but actually don't.
>>>> But if infinite sets do exist, then the set ℕ does exist, and all its
>>>> elements are members of finite intervals (0, n].
>>> No, any given element is a member of a finite set, but you can't then
>>> say that ALL are in such a set.
>> All are in the union of all finite sets.
> Why not just directly take N, made up of finite numbers?

Why not? Do it. Consider the black hats at every 10 n and white hats at 
all other numbers n. It is possible to shift the black hats such that 
every interval (0, n] is completely covered by black hats. There is no 
first n discernible that cannot be covered by  black hat. But the origin 
of each used black hat larger than n is now covered by a white hat. 
Without deleting all white hats it is not possible to cover all n by 
black hats. But deleting white hats is prohibited by logic. Exchanging 
can never delete one of the exchanged elements. Therefore we have here, 
like in all Cantor-pairings, the same impediment and further disussion 
is futile: You must deny logic. I do not.

Regards, WM